# 11.1: Chi-Square Test for Independence

Remember, qualitative data is where you collect data on individuals that are categories or names. Then you would count how many of the individuals had particular qualities. An example is that there is a theory that there is a relationship between breastfeeding and autism. To determine if there is a relationship, researchers could collect the time period that a mother breastfed her child and if that child was diagnosed with autism. Then you would have a table containing this information. Now you want to know if each cell is independent of each other cell. Remember, independence says that one event does not affect another event. Here it means that having autism is independent of being breastfed. What you really want is to see if they are not independent. In other words, does one affect the other? If you were to do a hypothesis test, this is your alternative hypothesis and the null hypothesis is that they are independent. There is a hypothesis test for this and it is called the Chi-Square Test for Independence. Technically it should be called the Chi-Square Test for Dependence, but for historical reasons it is known as the test for independence. Just as with previous hypothesis tests, all the steps are the same except for the assumptions and the test statistic.

Hypothesis Test for Chi-Square Test

1. State the null and alternative hypotheses and the level of significance
$$H_{o}$$: the two variables are independent (this means that the one variable is not affected by the other)
$$H_{A}$$: the two variables are dependent (this means that the one variable is affected by the other)
Also, state your $$\alpha$$ level here.
2. State and check the assumptions for the hypothesis test
1. A random sample is taken.
2. Expected frequencies for each cell are greater than or equal to 5 (The expected frequencies, E, will be calculated later, and this assumption means $$E \geq 5$$).
3. Find the test statistic and p-value
Finding the test statistic involves several steps. First the data is collected and counted, and then it is organized into a table (in a table each entry is called a cell). These values are known as the observed frequencies, which the symbol for an observed frequency is O. Each table is made up of rows and columns. Then each row is totaled to give a row total and each column is totaled to give a column total.
The null hypothesis is that the variables are independent. Using the multiplication rule for independent events you can calculate the probability of being one value of the first variable, A, and one value of the second variable, B (the probability of a particular cell $$P(A \text { and } B) )$$. Remember in a hypothesis test, you assume that $$H_{o}$$ is true, the two variables are assumed to be independent.

\begin{align*} P(A \text { and } B) &=P(A) \cdot P(B) \text { if } A \text { and } B are independent \\[4pt] &=\frac{\text { number of ways } A \text { can happen }}{\text { total number of individuals }} \cdot \frac{\text { number of ways } B \text { can happen }}{\text { total number of individuals }} \\[4pt] &= \frac{\text { row total }}{n} * \frac{\text { column total }}{n} \end{align*}

Now you want to find out how many individuals you expect to be in a certain cell. To find the expected frequencies, you just need to multiply the probability of that cell times the total number of individuals. Do not round the expected frequencies.

Expected frequency $$(\operatorname{cell} A \text { and } B)=E(A \text { and } B)$$

$$\begin{array}{l}{=n\left(\frac{\text { row total }}{n} \cdot \frac{\text { column total }}{n}\right)} \\ {=\frac{\text { row total -column total }}{n}}\end{array}$$

If the variables are independent the expected frequencies and the observed frequencies should be the same. The test statistic here will involve looking at the difference between the expected frequency and the observed frequency for each cell. Then you want to find the “total difference” of all of these differences. The larger the total, the smaller the chances that you could find that test statistic given that the assumption of independence is true. That means that the assumption of independence is not true. How do you find the test statistic? First find the differences between the observed and expected frequencies. Because some of these differences will be positive and some will be negative, you need to square these differences. These squares could be large just because the frequencies are large, you need to divide by the expected frequencies to scale them. Then finally add up all of these fractional values. This is the test statistic.

#### Test Statistic:

The symbol for Chi-Square is $$\chi^{2}$$

$$\chi^{2}=\sum \frac{(O-E)^{2}}{E}$$

where O is the observed frequency and E is the expected frequency

#### Distribution of Chi-Square

$$\chi^{2}$$ has different curves depending on the degrees of freedom. It is skewed to the right for small degrees of freedom and gets more symmetric as the degrees of freedom increases (see Figure 11.1.1). Since the test statistic involves squaring the differences, the test statistics are all positive. A chi-squared test for independence is always right tailed.

Figure 11.1.1: Chi-Square Distribution

p-value:

Using the TI-83/84: $$\chi \text { cdf (lower limit, } 1 \mathrm{E} 99, d f )$$

Using R: $$1-\text { pchisq }\left(x^{2}, d f\right)$$

Where the degrees of freedom is $$d f=(\# \text { of rows }-1) *(\# \text { of columns }-1)$$

4. Conclusion

This is where you write reject $$H_{o}$$ or fail to reject $$H_{o}$$. The rule is: if the p-value < $$\alpha$$, then reject $$H_{o}$$. If the p-value $$\geq \alpha$$, then fail to reject $$H_{o}$$.

5. Interpretation

This is where you interpret in real world terms the conclusion to the test. The conclusion for a hypothesis test is that you either have enough evidence to show $$H_{A}$$ is true, or you do not have enough evidence to show $$H_{A}$$ is true.

Example $$\PageIndex{1}$$ hypothesis test with chi-square test using formula

Is there a relationship between autism and breastfeeding? To determine if there is, a researcher asked mothers of autistic and non-autistic children to say what time period they breastfed their children. The data is in table #11.1.1 (Schultz, Klonoff-Cohen, Wingard, Askhoomoff, Macera, Ji & Bacher, 2006). Do the data provide enough evidence to show that that breastfeeding and autism are independent? Test at the1% level.

 Autis261m Breast Feeding Timelines Row Total None Less than 2 months 2 to 6 months More than 6 months Yes 241 198 164 215 818 No 20 25 27 44 116 Column Total 261 223 191 259 934

Table 11.1.1: Autism Versus Breastfeeding

Solution:

1. State the null and alternative hypotheses and the level of significance

$$H_{o}$$: Breastfeeding and autism are independent

$$H_{A}$$: Breastfeeding and autism are dependent

$$\alpha$$ = 0.01

2. State and check the assumptions for the hypothesis test

1. A random sample of breastfeeding time frames and autism incidence was taken.
2. Expected frequencies for each cell are greater than or equal to 5 (ie. $$E \geq 5$$). See step 3. All expected frequencies are more than 5.

3. Find the test statistic and p-value

Test statistic:

First find the expected frequencies for each cell

$$E(\text { Autism and no breastfeeding })=\frac{818^{*} 261}{934} \approx 228.585$$

$$E(\text { Autism and }<2 \text { months })=\frac{818^{*} 223}{934} \approx 195.304$$

$$E(\text { Autism and } 2 \text { to } 6 \text { months })=\frac{818^{*} 191}{934} \approx 167.278$$

$$E(\text { Autism and more than } 6 \text { months })=\frac{818 * 259}{934} \approx 226.833$$

Others are done similarly. It is easier to do the calculations for the test statistic with a table, the others are in table #11.1.2 along with the calculation for the test statistic. (Note: the column of O-E should add to 0 or close to 0.)

O E O-E $$(O-E)^{2}$$ $$(O-E)^{2} / E$$
241 228.585 12.415 154.132225 0.674288448
198 195.304 2.696 7.268416 0.03721591
164 167.278 -3.278 10.745284 0.064236086
215 226.833 -11.833 140.019889 0.617281828
20 32.4154 -12.4154 154.1421572 4.755213792
25 27.6959 -2.6959 7.26787681 0.262417066
27 23.7216 3.2784 10.74790656 0.453085229
44 32.167 11.833 140.019889 4.352904809
Total   0.0001   11.2166432 = $$\chi^{2}$$

Table 11.1.2: Calculations for Chi-Square Test Statistic

The test statistic formula is $$\chi^{2}=\sum \frac{(O-E)^{2}}{E}$$, which is the total of the last column in Table 11.1.2.

p-value:

$$d f=(2-1)^{*}(4-1)=3$$

Using TI-83/84: $$\chi \operatorname{cdf}(11.2166432,1 \mathrm{E} 99,3) \approx 0.01061$$

Using R: $$1-\text{pchisq}(11.2166432,3) \approx 0.01061566$$

4. Conclusion

Fail to reject $$H_{o}$$ since the p-value is more than 0.01.

5. Interpretation

There is not enough evidence to show that breastfeeding and autism are dependent. This means that you cannot say that the whether a child is breastfed or not will indicate if that the child will be diagnosed with autism.

Example $$\PageIndex{2}$$ hypothesis test with chi-square test using technology

Is there a relationship between autism and breastfeeding? To determine if there is, a researcher asked mothers of autistic and non-autistic children to say what time period they breastfed their children. The data is in Table 11.1.1 (Schultz, Klonoff-Cohen, Wingard, Askhoomoff, Macera, Ji & Bacher, 2006). Do the data provide enough evidence to show that that breastfeeding and autism are independent? Test at the1% level.

Solution:

1. State the null and alternative hypotheses and the level of significance

$$H_{o}$$: Breastfeeding and autism are independent

$$H_{A}$$: Breastfeeding and autism are dependent

$$\alpha$$ = 0.01

2. State and check the assumptions for the hypothesis test

1. A random sample of breastfeeding time frames and autism incidence was taken.
2. Expected frequencies for each cell are greater than or equal to 5 (ie. $$E \geq 5$$). See step 3. All expected frequencies are more than 5.

3. Find the test statistic and p-value

Test statistic:

To use the TI-83/84 calculator to compute the test statistic, you must first put the data into the calculator. However, this process is different than for other hypothesis tests. You need to put the data in as a matrix instead of in the list. Go into the MATRX menu then move over to EDIT and choose 1:[A]. This will allow you to type the table into the calculator. Figure 11.1.2 shows what you will see on your calculator when you choose 1:[A] from the EDIT menu.

Figure 11.1.2: Matrix Edit Menu on TI-83/84

The table has 2 rows and 4 columns (don’t include the row total column and the column total row in your count). You need to tell the calculator that you have a 2 by 4. The 1 X1 (you might have another size in your matrix, but it doesn’t matter because you will change it) on the calculator is the size of the matrix. So type 2 ENTER and 4 ENTER and the calculator will make a matrix of the correct size. See Figure 11.1.3.

Figure 11.1.3: Matrix Setup for Table

Now type the table in by pressing ENTER after each cell value. Figure 11.1.4 contains the complete table typed in. Once you have the data in, press QUIT.

Figure 11.1.4: Data Typed into Matrix

To run the test on the calculator, go into STAT, then move over to TEST and choose $$\chi^{2}$$-Test from the list. The setup for the test is in Figure 11.1.5.

Figure 11.1.5: Setup for Chi-Square Test on TI-83/84

Once you press ENTER on Calculate you will see the results in Figure 11.1.6.

Figure 11.1.6: Results for Chi-Square Test on TI-83/84

The test statistic is $$\chi^{2} \approx 11.2167$$ and the p-value is $$p \approx 0.01061$$. Notice that the calculator calculates the expected values for you and places them in matrix B. To eview the expected values, go into MATRX and choose 2:[B]. Figure 11.1.7 shows the output. Press the right arrows to see the entire matrix.

Figure 11.1.7: Expected Frequency for Chi-Square Test on TI-83/84

To compute the test statistic and p-value with R,
row1 = c(data from row 1 separated by commas)
row2 = c(data from row 2 separated by commas)
keep going until you have all of your rows typed in.
data.table = rbind(row1, row2, …) – makes the data into a table. You can call it what ever you want. It does not have to be data.table.
data.table – use if you want to look at the table
chisq.test(data.table) – calculates the chi-squared test for independence
chisq.test(data.table)$expected – let’s you see the expected values For this example, the commands would be row1 = c(241, 198, 164, 215) row2 = c(20, 25, 27, 44) data.table = rbind(row1, row2) data.table Output: [,1] [,2] [,3] [,4] row1 241 198 164 215 row2 20 25 27 44 chisq.test(data.table) Output: Pearson's Chi-squared test data: data.table X-squared = 11.217, df = 3, p-value = 0.01061 chisq.test(data.table)$expected

Output: [,1] [,2] [,3] [,4]
row1 228.58458 195.30407 167.27837 226.83298
row2 32.41542 27.69593 23.72163 32.16702

The test statistic is $$\chi^{2} \approx 11.217$$ and the p-value is $$p \approx 0.01061$$.

4. Conclusion

Fail to reject $$H_{o}$$ since the p-value is more than 0.01.

5. Interpretation

There is not enough evidence to show that breastfeeding and autism are dependent. This means that you cannot say that the whether a child is breastfed or not will indicate if that the child will be diagnosed with autism.

Example $$\PageIndex{3}$$ hypothesis test with chi-square test using formula

The World Health Organization (WHO) keeps track of how many incidents of leprosy there are in the world. Using the WHO regions and the World Banks income groups, one can ask if an income level and a WHO region are dependent on each other in terms of predicting where the disease is. Data on leprosy cases in different countries was collected for the year 2011 and a summary is presented in Table 11.1.3 ("Leprosy: Number of," 2013). Is there evidence to show that income level and WHO region are independent when dealing with the disease of leprosy? Test at the 5% level.

 WHO Region World Bank Income Group Row Total High Income Upper Middle Income Lower Middle Income Low Income Americas 174 36028 615 0 36817 Eastern Mediterranean 54 6 1883 604 2547 Europe 10 0 0 0 10 Western Pacific 26 216 3689 1155 5086 Africa 0 39 1986 15928 17953 South-East Asia 0 0 149896 10236 160132 Column Total 264 36289 158069 27923 222545

Table 11.1.3: Number of Leprosy Cases

Solution:

1. State the null and alternative hypotheses and the level of significance

$$H_{o}$$: WHO region and Income Level when dealing with the disease of leprosy are independent

$$H_{A}$$: WHO region and Income Level when dealing with the disease of leprosy are dependent

$$\alpha$$ = 0.05

2. State and check the assumptions for the hypothesis test

1. A random sample of incidence of leprosy was taken from different countries and the income level and WHO region was taken.
2. Expected frequencies for each cell are greater than or equal to 5 (ie. $$E \geq 5$$). See step 3. There are actually 4 expected frequencies that are less than 5, and the results of the test may not be valid. If you look at the expected frequencies you will notice that they are all in Europe. This is because Europe didn’t have many cases in 2011.

3. Find the test statistic and p-value

Test statistic:

First find the expected frequencies for each cell.

$$E(\text { Americas and High Income })=\frac{36817 * 264}{222545} \approx 43.675$$

$$E(\text { Americas and Upper Middle Income })=\frac{36817 * 36289}{222545} \approx 6003.514$$

$$E (\text { Americas and Lower Middle Income) }=\frac{36817 * 158069}{222545} \approx 26150.335$$

$$E(\text { Americas and Lower Income })=\frac{36817 * 27923}{222545} \approx 4619.475$$

Others are done similarly. It is easier to do the calculations for the test statistic with a table, and the others are in Table 11.1.4 along with the calculation for the test statistic.

O E O-E $$(O-E)^{2}$$ $$(O-E)^{2} / E$$
174 43.675 130.325 16984.564 388.8838719
54 3.021 50.979 2598.813 860.1218328
10 0.012 9.988 99.763 8409.746711
26 6.033 19.967 398.665 66.07628214
0 21.297 -21.297 453.572 21.29722977
0 189.961 -189.961 36085.143 189.9608978
36028 6003.514 30024.486 901469735.315 150157.0038
6 415.323 -409.323 167545.414 403.4097962
0 1.631 -1.631 2.659 1.6306365
216 829.342 -613.342 376188.071 453.5983897
39 2927.482 -2888.482 8343326.585 2850.001268
0 26111.708 -26111.708 681821316.065 26111.70841
615 26150.335 -25535.335 652053349.724 24934.7988
1883 1809.080 73.290 5464.144 3.020398811
0 7.103 -7.103 50.450 7.1027882
3689 3612.478 76.522 5855.604 1.620938405
1986 12751.636 -10765.636 115898911.071 9088.944681
149896 113738.368 36157.632 1307374351.380 11494.57632
0 4619.475 -4619.475 21339550.402 4619.475122
604 319.575 284.425 80897.421 253.1404187
0 1.255 -1.255 1.574 1.25471253
1155 638.147 516.853 267137.238 418.6140882
15928 2252.585 13675.415 187016964.340 83023.25138
10236 20091.963 -9855.963 97140000.472 4834.769106
Total   0.000   328594.008 = $$\chi^{2}$$

Table 11.1.4: Calculations for Chi-Square Test Statistic

The test statistic formula is $$\chi^{2}=\sum \frac{(O-E)^{2}}{E}$$, which is the total of the last column in Table 11.1.2.

p-value:

$$d f=(6-1) *(4-1)=15$$

Using the TI-83/84: $$\chi \operatorname{cdf}(328594.008,1 \mathrm{E} 99,15) \approx 0$$

Using R: $$1-\text { pchisq }(328594.008,15) \approx 0$$

4. Conclusion

Reject $$H_{o}$$ since the p-value is less than 0.05.

5. Interpretation

There is enough evidence to show that WHO region and income level are dependent when dealing with the disease of leprosy. WHO can decide how to focus their efforts based on region and income level. Do remember though that the results may not be valid due to the expected frequencies not all be more than 5.

Example $$\PageIndex{4}$$ hypothesis test with chi-square test using technology

The World Health Organization (WHO) keeps track of how many incidents of leprosy there are in the world. Using the WHO regions and the World Banks income groups, one can ask if an income level and a WHO region are dependent on each other in terms of predicting where the disease is. Data on leprosy cases in different countries was collected for the year 2011 and a summary is presented in Table 11.1.3 ("Leprosy: Number of," 2013). Is there evidence to show that income level and WHO region are independent when dealing with the disease of leprosy? Test at the 5% level.

Solution:

1. State the null and alternative hypotheses and the level of significance
$$H_{o}$$: WHO region and Income Level when dealing with the disease of leprosy are independent

$$H_{A}$$: WHO region and Income Level when dealing with the disease of leprosy are dependent

$$\alpha$$ = 0.05

2. State and check the assumptions for the hypothesis test

1. A random sample of incidence of leprosy was taken from different countries and the income level and WHO region was taken.
2. Expected frequencies for each cell are greater than or equal to 5 (ie. $$E \geq 5$$). See step 3. There are actually 4 expected frequencies that are less than 5, and the results of the test may not be valid. If you look at the expected frequencies you will notice that they are all in Europe. This is because Europe didn’t have many cases in 2011.

3. Find the test statistic and p-value
Test statistic:
Using the TI-83/84. See Example 11.1.2 for the process of doing the test on the calculator. Remember, you need to put the data in as a matrix instead of in the list.

Figure 11.1.8: Setup for Matrix on TI-83/84

Figure 11.1.9: Results for Chi-Square Test on TI-83/84

$$\chi^{2} \approx 328594.0079$$

Figure 11.1.10: Expected Frequency for Chi-Square Test on TI-83/84

Press the right arrow to look at the other expected frequencies.

p-value:

$$p-\text {value} \approx 0$$

Using R:
row1=c(174, 36028, 615, 0)
row2=c(54, 6, 1883, 604)
row3=c(10, 0, 0, 0)
row4=c(26, 216, 3689, 1155)
row5=c(0, 39, 1986, 15928)
row6=c(0, 0, 149896, 10236)
chisq.test(data.table)

Pearson's Chi-squared test

data: data.table
X-squared = 328590, df = 15, p-value < 2.2e-16

Warning message:
In chisq.test(data.table) : Chi-squared approximation may be incorrect

chisq.test(data.table)\$expected

$$\begin{array} {ccccc} {}&{[,1]}&{[,2]}&{[,3]}&{[,4]} \\ {\text{row1}}&{43.67515783}&{6003.514404}&{2.615034e+04}&{4619.475122}\\{\text{row2}}&{3.02144735}&{415.323117}&{1.809080e+03}&{319.575281}\\ {\text{row3}}&{0.01186277}&{1.630637}&{7.102788e+00}&{1.254713}\\{\text{row4}}&{6.03340448}&{829.341724}&{3.612478e+03}&{638.146793}\\{\text{row5}}&{21.29722977}&{2927.481709}&{1.275164e+04}&{2252.585405}\\{\text{row6}}&{189.96089780}&{26111.708410}&{1.137384e+05}&{20091.962686} \end{array}$$

Warning message:
In chisq.test(data.table) : Chi-squared approximation may be incorrect

$$\chi^{2}=328590$$ and p-value = $$2.2 \times 10^{-16}$$

4. Conclusion

Reject $$H_{o}$$ since the p-value is less than 0.05.

5. Interpretation

There is enough evidence to show that WHO region and income level are dependent when dealing with the disease of leprosy. WHO can decide how to focus their efforts based on region and income level. Do remember though that the results may not be valid due to the expected frequencies not all be more than 5.

#### Homework

Exercise $$\PageIndex{1}$$

In each problem show all steps of the hypothesis test. If some of the assumptions are not met, note that the results of the test may not be correct and then continue the process of the hypothesis test.

1. The number of people who survived the Titanic based on class and sex is in Table 11.1.5 ("Encyclopedia Titanica," 2013). Is there enough evidence to show that the class and the sex of a person who survived the Titanic are independent? Test at the 5% level.
 Class Sex Total Female Male 1st 134 59 193 2nd 94 25 119 3rd 80 58 138 Total 308 142 450

Table 11.1.5: Surviving the Titanic

2. Researchers watched groups of dolphins off the coast of Ireland in 1998 to determine what activities the dolphins partake in at certain times of the day ("Activities of dolphin," 2013). The numbers in Table 11.1.6 represent the number of groups of dolphins that were partaking in an activity at certain times of days. Is there enough evidence to show that the activity and the time period are independent for dolphins? Test at the 1% level.

 Activity Period Row Total Morning Noon Afternoon Evening Travel 6 6 14 13 39 Feed 28 4 0 56 88 Social 38 5 9 10 62 Column Total 72 15 23 79 189

Table 11.1.6: Dolphin Activity

3. Is there a relationship between autism and what an infant is fed? To determine if there is, a researcher asked mothers of autistic and non-autistic children to say what they fed their infant. The data is in Table 11.1.7 (Schultz, Klonoff-Cohen, Wingard, Askhoomoff, Macera, Ji & Bacher, 2006). Do the data provide enough evidence to show that that what an infant is fed and autism are independent? Test at the 1% level.

 Autism Feeding Row Total Breast feeding Formula with DHA/ARA Formula without DRA/ARA Yes 12 39 65 116 No 6 22 10 38 Column Total 18 61 75 164

Table 11.1.7: Autism Versus Breastfeeding

4. A person’s educational attainment and age group was collected by the U.S. Census Bureau in 1984 to see if age group and educational attainment are related. The counts in thousands are in Table 11.1.8 ("Education by age," 2013). Do the data show that educational attainment and age are independent? Test at the 5% level.

 Education Age Group Row Total 25-34 35-44 45-54 55-64 >64 Did not complete HS 5416 5030 5777 7606 13746 37575 Completed HS 16431 1855 9435 8795 7558 44074 College 1-3 years 8555 5576 3124 2524 2503 22282 College 4 or more years 9771 7596 3904 3109 2483 26863 Column Total 40173 20057 22240 22034 26290 130794

Table 11.1.8: Educational Attainment and Age Group

5. Students at multiple grade schools were asked what their personal goal (get good grades, be popular, be good at sports) was and how important good grades were to them (1 very important and 4 least important). The data is in Table 11.1.9 ("Popular kids datafile," 2013). Do the data provide enough evidence to show that goal attainment and importance of grades are independent? Test at the 5% level.

 Goal Grades Importance Rating Row Total 1 2 3 4 Grades 70 66 55 56 247 Popular 14 33 45 49 141 Sports 10 24 33 23 90 Column Total 94 123 133 128 478

Table 11.1.9: Personal Goal and Importance of Grades

6. Students at multiple grade schools were asked what their personal goal (get good grades, be popular, be good at sports) was and how important being good at sports were to them (1 very important and 4 least important). The data is in Table 11.1.10 ("Popular kids datafile," 2013). Do the data provide enough evidence to show that goal attainment and importance of sports are independent? Test at the 5% level.

 Goal Sports Importance Rating Row Total 1 2 3 4 Grades 83 81 55 28 247 Popular 32 49 43 17 141 Sports 50 24 14 2 90 Column Total 165 154 112 47 478

Table 11.1.10: Personal Goal and Importance of Sports

7. Students at multiple grade schools were asked what their personal goal (get good grades, be popular, be good at sports) was and how important having good looks were to them (1 very important and 4 least important). The data is in Table 11.1.11 ("Popular kids datafile," 2013). Do the data provide enough evidence to show that goal attainment and importance of looks are independent? Test at the 5% level.

 Goal Looks Importance Rating Row Total 1 2 3 4 Grades 80 66 66 35 247 Popular 81 30 18 12 141 Sports 24 30 17 19 90 Column Total 185 126 101 66 478

Table 11.1.11: Personal Goal and Importance of Looks

8. Students at multiple grade schools were asked what their personal goal (get good grades, be popular, be good at sports) was and how important having money were to them (1 very important and 4 least important). The data is in Table 11.1.12 ("Popular kids datafile," 2013). Do the data provide enough evidence to show that goal attainment and importance of money are independent? Test at the 5% level.

 Goal Money Importance Rating Row Total 1 2 3 4 Grades 14 34 71 128 247 Popular 14 29 35 63 141 Sports 6 12 26 46 90 Column Total 34 75 132 237 478

Table 11.1.12: Personal Goal and Importance of Money